The fixed point of homeomorphism on the circle Let $f$ be a homeomorphism on the circle $S^1$ which preserve the orientation, if the set of fixed points of $f$ is infinite, does it imply $f=id$? 
 A: No. Fix one hemisphere of the circle and then in the other hemisphere apply a non-identity endpoint-preserving homeomorphism to the interval. The fixed hemisphere is an infinite subset but the homeomorphism on the circle is constructed to be non-identity.
A: No. Here is a counterexample. Parameterize $S^1$ by $\gamma: [-1,1] \to S^1$. Let $f: [-1,1] \to [-1,1]$ be defined by
$$
f(x) = \begin{cases}
x & \text{if } x \leq 0 \\
x^2 & \text{if } x \gt 0
\end{cases}
$$
The function $f$ is a homeomorphism from $[-1,1]$ to itself that has $[-1,0] \cup \{1\}$ as its set of fixed points. The parametrization $\gamma$ then lifts this to $S^1$.
A: As you requested: 
Lemma. Suppose that $f$ is a homeomorphism of finite order of the circle which has at least 3 fixed points. Then $f=id$, the identity map. 
Proof. Let $x, y, z$ be distinct fixed points of $f$. I will identify $S^1$ with the 1-point compactification of the real line (using the stereographic projection). Choose this identification in such a way that $x$ is identified with $\infty$ and $y$ is identified with $0$. Since $f(0)=0, f(z)=z$, it follows that $f$ has to be strictly increasing as a homeomorphism of the real line. Therefore, $f|R$ belongs to the group $Homeo_+(R)$. The latter is left-orderable (see my answer here). It is immediate that every left-orderable group is torsion-free: If $g>1$ then 
$$g^2>g>1,$$
....
$$
g^n>g^{n-1}>...>g>1.$$ 
Hence, $g^n\ne 1$ for every $n>0$. Therefore, in our case, $f=1$, the identity homeomorphsm. qed 
Note that this lemma is a special case of a more general theorem on fixed-point sets of order $p$ homeomorphisms of homology spheres (given by the so called Smith Theory). Here $p$ is prime. Such fixed point sets are $p$-homology manifolds and, in addition, $p$-homology spheres. For instance, it follows from the Smith Theory that if $f: S^n\to S^n$ is a periodic homeomorphism then either $f=id$ or the fixed point set of $f$ has empty interior. Another example: if $f: S^2\to S^2$ is a homeomorphism of finite order, $f\ne id$, then the fixed-point set of $f$ is either a topological circle, or empty or a pair of distinct points. Same for $f: S^3\to S^3$, except you also have to allow a topological sphere as the fixed point set.  
